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Unformatted text preview: Math 598 Feb 11, 2005 1 Geometry and Topology II Spring 2005, PSU Lecture Notes 6 2.5 The inverse function theorem Recall that if f : M N is a diffeomorphism, then df p is nonsingular at all p M (by the chain rule and the observation that f f- 1 is the identity function on M ). The main aim of this section is to prove a converse of this phenomenon: Theorem 1 (The Inverse Function Theorem) . Let f : M N be a smooth map, and dim( M ) = dim( N ) . Suppose that df p is nonsingular at some p M . Then f is a local diffeomorphism at p , i.e., there exists an open neighborhood U of p such that 1. f is one-to-one on U . 2. f ( U ) is open in N . 3. f- 1 : f ( U ) U is smooth. In particular, d ( f- 1 ) f ( p ) = ( df p )- 1 . A simple fact which is applied a number of times in the proof of the above theorem is Lemma 2. Let f : M N , and g : N L be diffeomorphisms, and set h := g f . If any two of the mappings f , g , h are diffeomorphisms, then so is the third. In particular, the above lemma implies Proposition 3. If Theorem 1 is true in the case of M = R n = N , then, it is true in general. 1 Last revised: September 30, 2009 1 Proof. Suppose that Theorem 1 is true in the case that M = R n = N , and let f : M N be a smooth map with df p nonsingular at some p M . By definition, there exist local charts ( U, ) of M and ( V, ) of N , centered at p and f ( p ) respectively, such that f := - 1 f is smooth. Since and are diffeomorphisms, d p and d f ( p ) are nonsingular. Consequently, by the chain rule, d f o is nonsingular, and is thus a local diffeomorphism. More explicitly, there exists open neighborhoods A and B of the origin o of R n such that f : A B is a diffeomorphism. Since : - 1 ( A ) A is also a diffeomorphism, it follows that f : - 1 ( A ) B is a diffeomorphism. But f = f . So f : - 1 ( A ) B is a diffeomorphism. Finally, since : - 1 ( B ) B is a diffeomorphism, it follows, by the above lemma, that f : - 1 ( A ) - 1 ( B ) is a diffeomorphism. So it remains to prove Theorem 1 in the case that M = R n = N . To this end we need the following fact. Recall that a metric space is said to be complete provided that every Cauchy sequence of that space converges. Lemma 4 (The contraction Lemma) . Let ( X,d ) be a complete metric space, and < 1 . Suppose that there exists mapping f : X X such that d ( f ( x 1 ) , ( x 2 )) d ( x 1 ,x 2 ) , for all x 1 , x 2 X . Then there exists a unique point x X such that f ( x ) = x ....
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